\(\int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx\) [52]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{4 a x^2}+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^{3/2}} \] Output:

-1/2*(c*x^4+b*x^3+a*x^2)^(1/2)/x^3-1/4*b*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^2+1 
/8*(-4*a*c+b^2)*arctanh(1/2*x*(b*x+2*a)/a^(1/2)/(c*x^4+b*x^3+a*x^2)^(1/2)) 
/a^(3/2)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=-\frac {\sqrt {x^2 (a+x (b+c x))} \left (\sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)}+\left (b^2-4 a c\right ) x^2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )\right )}{4 a^{3/2} x^3 \sqrt {a+x (b+c x)}} \] Input:

Integrate[Sqrt[a*x^2 + b*x^3 + c*x^4]/x^4,x]
 

Output:

-1/4*(Sqrt[x^2*(a + x*(b + c*x))]*(Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x 
)] + (b^2 - 4*a*c)*x^2*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a] 
]))/(a^(3/2)*x^3*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1967, 1998, 27, 1951, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx\)

\(\Big \downarrow \) 1967

\(\displaystyle \frac {1}{4} \int \frac {b+2 c x}{x \sqrt {c x^4+b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3}\)

\(\Big \downarrow \) 1998

\(\displaystyle \frac {1}{4} \left (-\frac {\int \frac {b^2-4 a c}{2 \sqrt {c x^4+b x^3+a x^2}}dx}{a}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^4+b x^3+a x^2}}dx}{2 a}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3}\)

\(\Big \downarrow \) 1951

\(\displaystyle \frac {1}{4} \left (\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 a-\frac {x^2 (2 a+b x)^2}{c x^4+b x^3+a x^2}}d\frac {x (2 a+b x)}{\sqrt {c x^4+b x^3+a x^2}}}{a}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{3/2}}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{a x^2}\right )-\frac {\sqrt {a x^2+b x^3+c x^4}}{2 x^3}\)

Input:

Int[Sqrt[a*x^2 + b*x^3 + c*x^4]/x^4,x]
 

Output:

-1/2*Sqrt[a*x^2 + b*x^3 + c*x^4]/x^3 + (-((b*Sqrt[a*x^2 + b*x^3 + c*x^4])/ 
(a*x^2)) + ((b^2 - 4*a*c)*ArcTanh[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + 
b*x^3 + c*x^4])])/(2*a^(3/2)))/4
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1951

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] : 
> Simp[-2/(n - 2)   Subst[Int[1/(4*a - x^2), x], x, x*((2*a + b*x^(n - 2))/ 
Sqrt[a*x^2 + b*x^n + c*x^r])], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r, 2* 
n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]
 

rule 1967

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^p/(m + p*q 
+ 1)), x] - Simp[(n - q)*(p/(m + p*q + 1))   Int[x^(m + n)*(b + 2*c*x^(n - 
q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p - 1), x], x] /; FreeQ[{a, b, c}, x] 
&& EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] & 
& IGtQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q + 1, -(n - q) + 
 1] && NeQ[m + p*q + 1, 0]
 

rule 1998

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ 
.)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[A*x^(m - q + 1)*((a*x^q + b 
*x^n + c*x^(2*n - q))^(p + 1)/(a*(m + p*q + 1))), x] + Simp[1/(a*(m + p*q + 
 1))   Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p 
 + 1) + 1) - A*c*(m + p*q + 2*(n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b 
*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - 
q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] 
 && RationalQ[m, p, q] && ((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - 
q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*q + 1, 0]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (b x +2 a \right )}{4 x^{2} a}-\frac {\left (4 a c -b^{2}\right ) \left (-\ln \left (2\right )+\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )\right )}{8 a^{\frac {3}{2}}}\)	\(81\)
risch	\(-\frac {\left (b x +2 a \right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{4 x^{3} a}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{8 a^{\frac {3}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(108\)
default	\(-\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (4 c \,a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) x^{2}+2 c \sqrt {c \,x^{2}+b x +a}\, b \,x^{3}-4 c \sqrt {c \,x^{2}+b x +a}\, a \,x^{2}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{2} x^{2}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b x +2 \sqrt {c \,x^{2}+b x +a}\, b^{2} x^{2}+4 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \right )}{8 x^{3} \sqrt {c \,x^{2}+b x +a}\, a^{2}}\)	\(207\)

Input:

int((c*x^4+b*x^3+a*x^2)^(1/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/4*(c*x^2+b*x+a)^(1/2)*(b*x+2*a)/x^2/a-1/8*(4*a*c-b^2)/a^(3/2)*(-ln(2)+l 
n((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x/a^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.98 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {a} x^{3} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{16 \, a^{2} x^{3}}, -\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b x + 2 \, a^{2}\right )}}{8 \, a^{2} x^{3}}\right ] \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="fricas")
 

Output:

[-1/16*((b^2 - 4*a*c)*sqrt(a)*x^3*log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8* 
a^2*x - 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(a))/x^3) + 4*sqrt(c 
*x^4 + b*x^3 + a*x^2)*(a*b*x + 2*a^2))/(a^2*x^3), -1/8*((b^2 - 4*a*c)*sqrt 
(-a)*x^3*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(-a)/(a*c* 
x^3 + a*b*x^2 + a^2*x)) + 2*sqrt(c*x^4 + b*x^3 + a*x^2)*(a*b*x + 2*a^2))/( 
a^2*x^3)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}{x^{4}}\, dx \] Input:

integrate((c*x**4+b*x**3+a*x**2)**(1/2)/x**4,x)
 

Output:

Integral(sqrt(x**2*(a + b*x + c*x**2))/x**4, x)
 

Maxima [F]

\[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{3} + a x^{2}}}{x^{4}} \,d x } \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^4 + b*x^3 + a*x^2)/x^4, x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=\text {Timed out} \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^4,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^3+a\,x^2}}{x^4} \,d x \] Input:

int((a*x^2 + b*x^3 + c*x^4)^(1/2)/x^4,x)
 

Output:

int((a*x^2 + b*x^3 + c*x^4)^(1/2)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^4} \, dx=\frac {-4 \sqrt {c \,x^{2}+b x +a}\, a^{2}-2 \sqrt {c \,x^{2}+b x +a}\, a b x +4 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a c \,x^{2}-\sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b^{2} x^{2}-4 \sqrt {a}\, \mathrm {log}\left (x \right ) a c \,x^{2}+\sqrt {a}\, \mathrm {log}\left (x \right ) b^{2} x^{2}}{8 a^{2} x^{2}} \] Input:

int((c*x^4+b*x^3+a*x^2)^(1/2)/x^4,x)
 

Output:

( - 4*sqrt(a + b*x + c*x**2)*a**2 - 2*sqrt(a + b*x + c*x**2)*a*b*x + 4*sqr 
t(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a*c*x**2 - sqrt(a)* 
log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b**2*x**2 - 4*sqrt(a)*lo 
g(x)*a*c*x**2 + sqrt(a)*log(x)*b**2*x**2)/(8*a**2*x**2)